# If the Taylor Series of \ln(x) is known: \ln(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-... Can one find

If the Taylor Series of $\mathrm{ln}\left(x\right)$ is known:
$\mathrm{ln}\left(x\right)=\left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^{2}+\frac{1}{3}{\left(x-1\right)}^{2}-\frac{1}{4}{\left(x-1\right)}^{4}+\frac{1}{5}{\left(x-1\right)}^{5}-\dots$
Can one find the Taylor series of
$f\left(x\right)=\frac{x}{1-{x}^{2}}$
by manipulating the Taylor series of $\mathrm{ln}\left(x\right)$ ?
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Joseph Lewis
Make the substitution $x=1-{u}^{2}$ in order to obatin:
$\mathrm{ln}\left(1-{u}^{2}\right)=-{u}^{2}-\frac{{u}^{4}}{2}-\frac{{u}^{6}}{3}-\dots =-\sum _{k=1}^{\mathrm{\infty }}\frac{{u}^{2k}}{k}$
Based on the ratio test, the corresponding function is analytic for $|u|<1$. Prescisely,
$\underset{k\to \mathrm{\infty }}{lim}|\frac{{a}_{k+1}}{{a}_{k}}|=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{k}{k+1}\right)|{u}^{2}|=|{u}^{2}|<1\to |u|<1$
Hence we can differentiate the obtained expression in order to get:
$-\frac{2u}{1-{u}^{2}}=-2\sum _{k=1}^{\mathrm{\infty }}{u}^{2k-1}\to f\left(x\right)=\frac{x}{1-{x}^{2}}=\sum _{k=1}^{\mathrm{\infty }}{x}^{2k-1}=x+{x}^{3}+{x}^{5}+\dots$
whenever $|x|<1$, and we are done.
Hopefully this helps!
###### Not exactly what you’re looking for?
Paul Mitchell
user_27qwe

Notice that:
$\int f\left(x\right)dx=\int \frac{x}{1-{x}^{2}}dx=-\frac{1}{2}\int \frac{du}{u}=-\frac{1}{2}\mathrm{ln}|1-{x}^{2}|+C$
(via the u-substitution $u=1-{x}^{2}$)
$\mathrm{ln}\left(x\right)=\left(x-1\right)-\frac{1}{2}\left(x-1{\right)}^{2}+\frac{1}{3}\left(x-1{\right)}^{2}+...$
So substitute $1-{x}^{2}$ into x in the natural log taylor expansion, and take the derivative to obtain a series representation of f(x).